Quiz

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As an alternative to Coulomb's law, Gauss' law can be used to determine the electric field of charge distributions with symmetry. Integration of the electric field then gives the capacitance of conducting plates with the corresponding geometry.

Introduction

where \( Q_\text{encl} \) is the total charge enclosed within \( S \). If a distribution displays some symmetry (i.e., \( \mathbf{E} \) depends on \( d\mathbf{A} \) in a simple way or not at all), then the surface integral need not be computed directly.

The standard examples for which Gauss' law is often applied are spherical conductors, parallel-plate capacitors, and coaxial cylinders, although there are many other neat and interesting charges configurations as well.

To compute the capacitance, first use Gauss' law to compute the electric field as a function of charge and position. Next, integrate to find the potential difference, and, lastly, apply the relationship \( C = Q/\Delta V \).

Calculation of the electric field

The following examples illustrate the elementary use of Gauss' law to calculate the electric field of various symmetric charge configurations.

For \( r < R \), no charge is enclosed in the Gaussian surface, so the field is also zero.

Infinite plane with charge. An infinite plane of charge has uniform surface charge density \( \sigma \). Determine the electric field due to the plane.

Choose as a Gaussian surface a cylinder (or prism) whose faces are parallel to the sheet, each a distance \( r \) from the sheet. By symmetry, the electric field must point perpendicular to the plane, so the electric flux through the sides of the cylinder must be zero. If the area of each face is \( A \), then Gauss' law gives

\[ 2 A E = \frac{A\sigma}{\epsilon_0}, \]

so

\[ E = \frac{\sigma}{2\epsilon_0}. \]

Note that \( E \) is constant and independent of \( r \). This can be justified by considering the fact that the problem looks "identical" no matter how far one "zooms out"; an infinite plane is infinite. (And in fact, this argument is sufficient establish that \( E \) is a constant.)

Charged cylinder. An hollow cylindrical rod of radius \( R \) has uniform charge per unit length \( \lambda \). Determine the electric field due to the rod.

We choose as our Gaussian surface a concentric cylinder of radius \( r > R \). Consider a segment of rod of length \( L \). By symmetry, the electric field must point radially outward, so outside of the rod, Gauss' law gives

\[ E (2\pi r L) = \frac{\lambda L}{\epsilon_0}, \]

so

\[ E = \frac{\lambda}{2\pi \epsilon_0 r}. \]

Inside the rod, no charge is enclosed, so the flux through a concentric cylindrical Gaussian surface of radius \( r < R \) is zero, and therefore the electric field inside the rod is zero.

Sphere with hole. A hollow charged sphere of radius \( R \) and surface charge density \( \sigma \) contains a small circular hole of radius \( r \ll R \). What is the electric field right inside and outside of the hole?

Arbitrarily close to the hole, the hole essentially looks like a missing infinite plane. Thus, we can consider the electric field as resulting from the superposition of an infinite plane of charge density \( -\sigma \) and a charged sphere of charge density \( +\sigma \).

Right outside of the hole, the field due to the plane is \( \sigma / (2 \epsilon_0) \) inward while the field due to the sphere is \( 4 \pi R^2 \sigma / (4 \pi \epsilon_0 R^2) = \sigma / \epsilon_0 \) outward, so the net field is \( \sigma / (2 \epsilon_0) \) outward.

Right inside the hole, the field due to the plane is \( \sigma / (2 \epsilon_0) \) outward while the field due to the sphere is zero, so the net field is again \( \sigma / (2 \epsilon_0) \) outward.

We have a triangular uniformly charged plate of charge density \(\sigma\).

We take a point just above a vertex of the triangular plate at a distance \(d\) perpendicular to the plane of the triangle. Let the vertical component of electric field at that point be \(E(d)\) in \(V/m\)

Two infinitely large metal sheets have surface charge densities \( + \sigma \) and \( - \sigma, \) respectively. If they are kept parallel to each other at a small separation distance of \( d, \) what is the electric field at any point in the region between the two sheets?

Use \( \varepsilon_{0} \) for the permittivity of free space.

Calculation of the Capacitance

The following examples illustrate how to calculate the capacitance of some of the most frequently encountered systems.

Parallel-plate capacitor. Two parallel identical conducting plates, each of area \( A \), are separated by a distance \( d \). Determine the capacitance of the plates.

Let the plates be aligned with the \( xy \) plane, and suppose the bottom plate holds charge \( Q \) while the other holds charge \( -Q \). The field between the plates is the sum of the contribution from each plate (which point in the same direction between the plates), which we know to be \( E = \frac{\sigma}{2\epsilon_0} = \frac{Q}{2\epsilon_0 A} \) upward. Integrating \( 2E \) over the separation distance \( d \) gives

\[ \Delta V = -\int_0^d 2E \, dr = -\frac{Qd}{\epsilon_0 A}. \]

It follows that

\[ C = \frac{Q}{|\Delta V|} = \frac{\epsilon_0 A}{d}. \]

Note that the field in the region outside of the plates is zero.

Capacitance of a spherical capacitor. Determine the capacitance of a conducting sphere of radius \( R \).

Using Gauss' law, it is easy to show that the electric field from a charged sphere is identical to that of a point source outside of the sphere. In other words, at a distance \( r \) from the center of the sphere,

\[ E(r) = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2}, \]

where \( Q \) is the net charge of the sphere. A single-sphere capacitor is essentially a double-sphere concentric capacitor with the outer sphere held at infinity. (Equivalently, one can simply consider the potential far away to be infinite.) Integrating gives

I want to make a parallel plate capacitor using circular discs having radius equal to radius of Earth(Not really!) and having capacitance equal to 1 F. Then what must be the distance between those two discs in metre ?

Assumption:

Radius of Earth = \( 6.4 \times 10^{6} \text{m} \)

\( \epsilon_{0} = 8.85 \times 10^{-12} \dfrac{F}{m} \)

27 small drops , having charge q and radius r coalesce to form a bigger drop.How many times capacitance will become?